You have found the following ages (in years) of all 5 tigers at your local zoo: $ 7,\enspace 5,\enspace 4,\enspace 17,\enspace 28$ What is the average age of the tigers at your zoo? What is the variance? You may round your answers to the nearest tenth.
Explanation: Because we have data for all 5 tigers at the zoo, we are able to calculate the population mean $({\mu})$ and population variance $({\sigma^2})$ To find the population mean , add up the values of all $5$ ages and divide by $5$ $ {\mu} = \dfrac{\sum\limits_{i=1}^{{N}} x_i}{{N}} = \dfrac{\sum\limits_{i=1}^{{5}} x_i}{{5}} $ $ {\mu} = \dfrac{7 + 5 + 4 + 17 + 28}{{5}} = {12.2\text{ years old}} $ Find the squared deviations from the mean for each tiger. Age $x_i$ Distance from the mean $(x_i - {\mu})$ $(x_i - {\mu})^2$ $7$ years $-5.2$ years $27.04$ years $^2$ $5$ years $-7.2$ years $51.84$ years $^2$ $4$ years $-8.2$ years $67.24$ years $^2$ $17$ years $4.8$ years $23.04$ years $^2$ $28$ years $15.8$ years $249.64$ years $^2$ Because we used the population mean $({\mu})$ to compute the squared deviations from the mean , we can find the variance $({\sigma^2})$ , without introducing any bias, by simply averaging the squared deviations from the mean $ {\sigma^2} = \dfrac{\sum\limits_{i=1}^{{N}} (x_i - {\mu})^2}{{N}} $ $ {\sigma^2} = \dfrac{{27.04} + {51.84} + {67.24} + {23.04} + {249.64}} {{5}} $ $ {\sigma^2} = \dfrac{{418.8}}{{5}} = {83.76\text{ years}^2} $ The average tiger at the zoo is 12.2 years old. The population variance is 83.76 years $^2$.